 Cubic elimination at a nilpotent double-zero point

For a vector field with linear part $\dot{u}=v$, $\dot{v}=0$ and cubic terms $a_i{u}^3+b_i{u}^{2}v+c_{i}u{v}^2+d_i{v}^3$, write $(u,v)=(x,y)+(h_1,h_2)$ with homogeneous cubic $h_i$. The transformed cubic field is $f_3+Ah-DhA(x,y{)}^T$, where $A(x,y{)}^T=(y,0{)}^T$. Taking the coefficients of $h_2$ to be $(-a_1,c_2/2,d_2,0)$ and those of $h_1$ to be $((b_1+c_2/2)/3,(c_1+d_2)/2,d_1,0)$ eliminates every cubic term in $\dot{x}$ and the $x{y}^2,y^3$ terms in $\dot{y}$. The remaining normal form is $\dot{x}=y$, $\dot{y}=P{x}^3+Q{x}^{2}y$ through cubic order. The invariant damping coefficient contains the contribution $3a_1$ from differentiating the first equation.